Question #b93a3
1 Answer
Explanation:
The problem provides you with the density of the salt because you must use it to find the mass of the sample.
You can use the conversion factors
#color(blue)(ul(color(black)("1 L" = "1 dm"^3)))" "# and#" " color(blue)(ul(color(black)("1 dm"^3 = 10^3"cm"^3)))#
to calculate the mass of
#0.473 color(red)(cancel(color(black)("L"))) * (1color(red)(cancel(color(black)("dm"^3))))/(1color(red)(cancel(color(black)("L")))) * (10^3color(red)(cancel(color(black)("cm"^3))))/(1color(red)(cancel(color(black)("dm"^3)))) * "2.66 g"/(1color(red)(cancel(color(black)("cm"^3)))) = 1.258 * 10^3 "g"#
To find the concentration of the salt, use the molar mass of magnesium sulfate to find the number of moles present in the sample
#1.258 * 10^3color(red)(cancel(color(black)("g"))) * "1 mole MgSO"_4/(120.366color(red)(cancel(color(black)("g")))) = "10.45 moles MgSO"_4#
Keep in mind that molarity requires liters of solution, so use the fact that
#color(blue)(ul(color(black)("1 m"^3 = 10^3"dm"^3#
to find the volume of the solution -- you can assume that the volume of the solution will not change upon the addition of the salt
#1.00 color(red)(cancel(color(black)("m"^3))) * (10^3color(red)(cancel(color(black)("dm"^3))))/(1color(red)(cancel(color(black)("m"^3)))) * "1 L"/(1color(red)(cancel(color(black)("dm"^3)))) = 1.00 * 10^3"L"#
This means that the concentration of the salt will be
#c_ ("MgSO"_4) = "10.45 moles"/(1.00 * 10^3"L") = color(darkgreen)(ul(color(black)("0.0105 mol L"^(-1))))#
The answer is rounded to three sig figs.
